Algebra

br JavaScript is required for your course . ensure JavaScript is enabled in your browser preferencesSeminar on Linear Equations , Inequalities and GraphingTo earn seminar credit for this unit , eff one of the following options belowOption 1 : Participate in a synchronous seminar (Note : Students atomic number 18 encouraged to flow seminar but the educatee may complete Option 2 instead of aid seminarThe seminar s is Linear Equations , Inequalities and Graphing complete the Unit 3 reading sooner attending the seminar session in to be familiar with the unavoidable terminology and concepts Concepts and example problems will be discussed , supplementing the material cover by the textbookOption 2 : Complete the wide problem amaze below1 . Find the mid gratuity of the track segment PQ for(2 , -2 , Q (3 , 4Solution : The mid po int can be calculate asx (2 3 /2 5 /2y (-2 4 /2 2 /2 1Hence , the midpoint is (5 /2 , 12 .

If M (6 , 1 ) is the midpoint of segment PQ and the coordinates of Q are (2 , -2 , find the coordinates ofp Solution : permit the coordinates of pointbe (x , yThen (x 2 /2 6 and (y (-2 /2 1Therefore , x 2 12 and y - 2 2Therefore , x 10 and y 4Hence the pointis (10 , 43 . Verify that (-1 , 3 ) is a solution to y (-1 /2 )x (5 /2Solution : Substituting x -1 and y 3 in the given equation yieldsRHS (-1 /2 )x (5 /2 (-1 /2 (-1 (5 /2 (1 /2 (5 /2 3 y LHSHence (-1 , 3 ) is a solution of the given equation4 . light up the difference and spell the solution set in detachment annotation5 . Solve the inequality and write the solution set in interval nota! tion SolutionGiven (3 /5 )x (2 /3 (x - 5 x FGHGYUHMAjtE /3 )x - (10 /3 xTherefore (3 /5 )x (2 /3 )x -x 10 /3Therefore (9 10 - 15 /15 )x 10 /3Therefore (4 /15 )x 10 /3Therefore , x (10 15 (4 3Hence , x 25 /2Hence , the solution set is (- , 25 /2...If you want to get along a full essay, order it on our website: OrderEssay.net

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